∫Calc Practice

Order and linearity of differential equations

Problem 6.734 · easy

Classify \( \displaystyle 2 y + y' + 4 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = -2 + e^{- 2 x} \) is a solution.
  1. The highest derivative is y′, so the order is 1.
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is nonzero: nonhomogeneous.
  3. \[ \frac{d}{d x} \left(-2 + e^{- 2 x}\right) + 2 e^{- 2 x} = 0 \]
    Substituting y = -2 + exp(-2*x) makes the left side 0.✓ Proved
Answer \( \text{order 1, linear and nonhomogeneous} \)

Lines: 1 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0order from sympy.ode_order; linearity from the equation's second partials in y, y′, y″; the solution checked in floating point

Reviewers

  • gpt-oss:20b: fail (error) — The verification step omits the terms 2y and 4 from the differential equation; it only checks y' + 2e^{-2x}=0, which is not equivalent to 2y + y' + 4 = 0. The conclusion that the proposed y satisfies the equation is therefore incorrect.
  • qwen3.6:27b-mlx: pass — The classification is correct and the verification step correctly substitutes the proposed solution into the differential equation.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-10-25 — The classification is correct and the verification step correctly substitutes the proposed solution into the differential equation.
  • gpt-oss:20b: fail (error) 2026-10-25 — The verification step omits the terms 2y and 4 from the differential equation; it only checks y' + 2e^{-2x}=0, which is not equivalent to 2y + y' + 4 = 0. The conclusion that the proposed y satisfies the equation is therefore incorrect.
  • qwen3.6:27b-mlx: pass 2026-10-24 — The classification is correct and the verification step correctly substitutes the proposed solution into the differential equation.
  • gpt-oss:20b: fail (error) 2026-10-24 — The verification step omits the 2y and constant terms; it only checks y' + 2e^{-2x}=0, which is not equivalent to the original differential equation 2y + y' + 4 = 0. The conclusion that the proposed y satisfies the equation is therefore unsupported.
  • qwen3.6:27b-mlx: fail (error) 2026-10-24 — The verification step is algebraically incorrect. Substituting y = -2 + e^{-2x} into 2y + y' + 4 yields 4 + 2e^{-2x} - 2e^{-2x} = 4, not 0. The equation provided in line 3 also omits the constant term from 2y and 4, failing to represent the original ODE.
  • gpt-oss:20b: fail (error) 2026-10-24 — The verification step omits the terms 2y and 4 from the differential equation, so it does not actually check that y = -2 + e^{-2x} satisfies 2y + y' + 4 = 0.

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/de_classify, checked 2026-10-25 with SymPy 1.14.0.